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How to find the radius of a circle by knowing two points and its arc length

  1. Jan 7, 2012 #1
    How can I find the radius of a circle by knowing two points and its arc length? Do I have to use a numerical method to solve for a trigonometric equation or is there any algebraic or geometric method?
  2. jcsd
  3. Jan 7, 2012 #2
    Orient the two points to lie on the horizontal (just makes description easier) and draw a line between them. The part of the arc that lies above halfway along this line is now a third point, draw a line vertically down from this point. Now draw a third line from one of your two original points to the new one drawn on the arc, the angle this new line makes with the vertical is the angle a fourth (and final) line needs to be drawn at from the other end of your newest line towards the vertical one to form an isosceles triangle with the two equal length sides being the radii.

    Attached is an image which I'm sure will be easier to follow

    Last edited: Jan 7, 2012
  4. Jan 7, 2012 #3
    If by the radii you mean the two segments marked with the orange lines, then no, that's not the radius. You can't just assume that the segment of arc that you picked forms half the circle.

    I was thinking if you could find the curvature of the circle somehow, then you'd be done. This is a problem seeing as finding the curvature of any curve usually require you to know at least parametric equations of the curve. Do you have any more information regarding these two points, are they arbitrarily spread out on the circle?
  5. Jan 7, 2012 #4

    Thank you for the reply but as you did not use the length of the arc it is impossible to reach a solution by your way. Actually there are number of mistakes but I am sure you can notice them at a second glance. Again thank you for your effort.

    Unfortunately, yes they are arbitrary.
  6. Jan 7, 2012 #5
    So just to clear things up, by "two points known" you mean the coordinates are known? And the arc length is the length of the arc contained between the two points correct? (There are actually two ways to see this, any two points on a circle cut the circle into two parts of possibly different arc lengths). In either case, I don't think you can find the radius of the circle with the given information (at least not that I see right away). I'll keep thinking about it.
  7. Jan 7, 2012 #6
    Yes, the coordinates of the two points and the length of the arc between them are known. The solution that I get contains trigonometric identities. I can write it if you are curious, but what I need is an equation that is solvable by elementary methods.
  8. Jan 7, 2012 #7

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    This problem is not solvable by elementary methods. You are inevitably going to end up with a transcendental expression.

    It is easily solvable by approximation techniques.
  9. Jan 7, 2012 #8
    I did so. Since the process must be repeated, that will be time consuming but as you said and I thought, there seems to be no other way. Thank you for the answer.
  10. Jan 7, 2012 #9
    No, the two angles marked [itex]\phi[/itex] form the base of an isosceles triangle who's equal sides are the radius, the two lines marked with orange lines are just of equal length.

    But the arc length has been used, [itex]\phi[/itex] will become smaller with a larger arc length and larger with a smaller one.
  11. Jan 8, 2012 #10

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    Try your technique with an arc length of 14.1897054604163922812851617102553083 and the two points being the origin and (14,0). There's no room in that picture for the radius of the circle (which happens to be 25 in this example).
  12. Jan 8, 2012 #11
    Maybe I didn't explain well enough because my method gets 25 no problem, taking your values I found the segment height (using a calculator online) to be 1, this gave me a base angle for my isosceles triangle of [itex]\tan^{-1} 7[/itex] and a base length of [itex]\sqrt{50}[/itex] which therefore gives the equal sides a length of 25 each.
  13. Jan 8, 2012 #12

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    Maybe you didn't.

    There's no escaping that solving this problem will require solving a transcendental equation of some form, and that is what uyger asked in the original post. (He just didn't know that that was what he was asking.)
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